**NCERT Class 9 Science Chapter 11 Work and Energy Extra Questions and Answers**

Class 9 Science Chapter 11 Extra Inside Questions and Answers – Work and Energy. Here in this Page Class IX Students can Learn Extra Questions & Answer 11th Chapter Science fully Inside.

We Provided Here Work and Energy Science Chapter 11 Long Answer Type Question, Multiple Choice Questions & Answer, Short Answer Type Questions (2 or 3 marks), and Very Short Answer Type Question (1 marks) Solution.

**Class 9 Science Chapter 11 Extra Question with Answer – Work and Energy**

Science Chapter 11 **Work and Energy **Class 9 Inside 5 Marks, 3 marks, 2 Marks & And 1 Marks Important Questions and Answers.

**Very Short Answer Type Questions:**

**1.) Amit put a wooden block on his shoulder for 10 minutes. Does work done in this example?**

Answer: No, work is not done in this example because the value of displacement in this example is zero.

**2.) A force of 2 N applied on an object but the object remains in stationary state. Calculate work done by applied force.**

Answer: The displacement of the particle is zero thus work done becomes zero.

**3.) State the most common example of source of energy.**

Answer: Sun is the example of most common source of energy.

**4.) State the two forms of mechanical energy.**

Answer: There are two types of energy-

I.) Potential energy.

II.) Kinetic energy

**5.) Give the SI unit of energy.**

Answer: SI unit of energy is Joule.

**6.) Convert 1 KJ into J.**

Answer: As we know that, kilo means 10³. Thus

1 KJ = 10³ J.

**7.) On what factors does the kinetic energy depend?**

Answer: Kinetic energy is depend on mass and velocity of the object.

**8.) On what factors does the gravitational potential energy depend?**

Answer: Gravitational potential energy is depend on mass and height of the object.

**9.) Explain the energy changes in the following statement.**

Water heater increases the temperature of water.

Answer: in this process electrical energy is converted into heat energy.

**10.) Which is the type of energy associated with stretched string.**

Answer: Stretched string possesses energy because of its shape so the energy related with is a potential energy.

**11.) The electron moves along a nucleus in circular path because of electrostatic force. Calculate the work done by electrostatic force.**

Answer: Electron perform circular motion and we know that, work done in a cyclic process is zero.

**12.) Identify the type of energy in following example.**

**An arrow is put on stretched string.**

Answer: The energy contained in the arrow because of its specific position thus it contained potential energy.

**13.) On what factor does the potential energy of a stretched spring depends?**

Answer: Potential energy of A stretched spring is depend on the displacement from mean position.

**14.) Give two units of power.**

Answer: Watt and horse power are the units of power.

**15.) A boy apply force of 40 N on a car but the car remains stationary. Calculate work done. **

Answer: The car does not move so displacement becomes zero. Thus the value of work done will be zero.

**Short Answer Type Questions:**

**1.) On what factors does the work done depend?**

Answer:

As we know that work done is calculated by following formula.

Work done = force × displacement.

So we conclude that work done is depend on following quantities-

I.) Applied force

II.) Displacement due to applied force.

III.) Direction of applied force and displacement.

**2.) An object of mass 80 kg is moving with velocity 20 m/s. Calculate Kinetic energy associated with it.**

Answer: Given,

Mass = 80 kg,

Velocity = 20 m/s.

As we know that,

Kinetic energy = ½ mv²

Kinetic energy = ½ × 80 × 20²

Kinetic energy = 40 × 400

Kinetic energy = 16000 J

Kinetic energy = 16 K J

The energy associated with that object is 26 KJ.

**3.) A weightlifter lifts a stone of 80 kg at 160 cm. Calculate work done in this process by weightlifter.**

Answer: Given,

Mass = 80 kg,

Height = 160 cm = 1.6 m

As we know that, weightlifter did work against gravitational force. Thus the energy associated with this is gravitational potential energy.

As,

Gravitational potential energy = mgh

Gravitational potential energy = 80 × 9.8 × 1.6

Gravitational potential energy = 1254.4 J

Weightlifter did work done of 1254.4 J.

**4.) State the similarities between work and energy.**

Similarities between work and energy.

I.) Work and energy both has unit. The unit of work and energy is Joule.

II.) Work and energy does not depend on path.

III.) Work and energy both are scalar.

IV.) Work done means change in energy.

**5.) A force of 70 N applied on a wooden block and it displaced 56 m. Calculate work done. **

Answer :

Given, force = 70 N,

Displacement = 56 m.

As we know that,

Work done = force × displacement

Work done = 70 × 56

Work done = 3920 J.

The work done due to applied force is 3920 J.

**6.) A stone of mass 3 kg is released from 90 m. Calculate Kinetic energy when the stone reaches at ground. (Take g =10 m/s²)**

Answer: Given,

Height = s = 90 m, Mass = 3 kg,

The stone is released, so initial velocity = 0 m/s.

g = 10 m/s²,

We must know the value of final velocity to calculate Kinetic energy.

According to 3rd equation of motion,

v² = u² +2as

v² = 0² + 2 × 10 × 90

v² = 1800 m/s.

As we know that,

Kinetic energy = ½ mv²

Kinetic energy = ½ × 3 × 1800

Kinetic energy= 2700 J.

The kinetic energy of that stone will be 2700 J.

**7.) State the four examples of potential energy.**

Answer: Examples of potential energy-

I.) Stretched string

II.) Compressed rubber band

III.) Stone is at some height

IV.) Water stored in dam

**8.) A heater consumes 120000 joules of energy in 1 minutes. Calculate power**

Answer: Given, energy = 120000 J,

Time = 1 minute = 60 seconds,

As we know that

Power= energy/time

Power = 120000/60

Power= 2000 Watt.

The power of heater is 2000 Watt.

**9.) A stone of mass 8 kg is released from some height and it reach at ground after 20 seconds. Calculate work done by gravitational force. (Take g = 10 m/s²)**

Answer: Given, The stone is released , so initial velocity = u = 0 m/s,

Time = 20 seconds, g = 10 m/s²,

Mass = 8 kg.

We calculate final velocity using 1st equation of motion.

As,

v = u + at

v = 0 + 10×20

v = 200 m/s.

As we know that change in kinetic energy is called as work done.

Work done = change in kinetic energy

Work done = final Kinetic energy- initial kinetic energy.

Work done = ½ mv² – ½ mu²

Work done = ½ × m(v² – u²)

Work done = ½ × 8 × (200²-0²)

Work done = 4 ×( 40000)

Work done = 160000 J

Work done = 160 KJ.

The work done by gravitational force is 160 KJ.

**10.) A motorcycle of mass 150 kg is moving with 54 km/hr. Calculate Kinetic energy associated with it. **

Answer: Given,

Mass = 150 kg,

Velocity = 54 km/hr.

As,

54 km/hr = 54 × 5/18 m/s

54 km/hr = 15 m/s.

Velocity = 15 m/s.

As we know that,

Kinetic energy = ½ mv²

Kinetic energy = ½ × 150 × 15²

Kinetic energy = 75 × 225

Kinetic energy = 16875 J.

Kinetic energy = 16.9 KJ

The kinetic energy associated with motorcycle is 16.9 KJ.

**11.) Explain the relationship between acceleration and kinetic energy.**

Answer:

We know that, Kinetic energy is depend on square of velocity.

If the velocity of the object increases then its kinetic energy also increases.

Also we know that, Acceleration means increases in velocity with time.

When we increases the velocity then kinetic energy of the object also increases.

Thus we say that, Kinetic energy of the object increases with acceleration.

**In case you are missed :- Previous Chapter Extra Questions**

**12.) The object of weight 90 N is moving with velocity 30 m/s. Calculate Kinetic energy associated with it. (Take g = 10 m/s²)**

Answer: Given,

Weight = 90 N, g = 10 m/s²,

Velocity = v = 30 m/s.

As we know,

Weight = mass × acceleration due to gravity

Mass = weight/ acceleration due to gravity

Mass = 90/10

Mass = 9 kg,.

As,

Kinetic energy = ½ mv²

Kinetic energy = ½ × 9 × 30²

Kinetic energy = 4.5 × 900

Kinetic energy = 4050 J.

The kinetic energy associated with the object is 4050 J.

**13.) A stone of mass 230 gm is falling from top of the tower of height 80 m. The resistance of air equal to gravitational force applying on the object. Calculate work done by net force.**

**Answer: **Given, Mass = 230 gm = 0.23 kg

The downward force is equal to upward air resistance so it fall with constant velocity. The net force on object is zero.

As we know,

Work done = force × displacement

Work done = 0 × 80

Work done= 0 J.

The work done by net force is zero.

**14.) A man did work of 5230 J in 1 minute. Calculate power. **

**Answer:**

Given, work done = 5230 J,

time = 1 minutes = 60 seconds,

As we know that,

Power = work done/time

Power = 5230/60

Power = 87.17 watt.

The value of power is 87.17 watt.

**15.) A heater consumes energy 220 KJ in 2 minutes. Calculate power rating of that heater.**

**Answer:**

Given, energy = 220 KJ = 220 ×10³ J = 220000

Time = 2 minutes = 120 seconds,

As we know that,

Power = energy/ time

Power = 220000/120

Power = 1833.3 watt.

The power rating of the heater is 1833.3 watt.

**Long Answer Type Questions**

**1.) a) Explain the concept of work done in your own words.**

**b) A car of mass 400 kg is moving with velocity 32 m/s. Suddenly driver saw an obstacle on road therefore he breaks the car and it stopped. Calculate work done.**

Answer: a)

**Work done-**

Work done is a quantity which depends on force and displacement.

If an object displaced because of force then we say that work is done It is mathematically expressed as,

*Work done = force × displacement *

Work done also depends on direction of applied force. If the direction of force and displacement is same then we say that the work done is positive. If the direction of force and displacement is opposite then we say that the work done is negative. If the angle between force and displacement is 90⁰ then work done becomes zero. We can express work done in term of energy also. Change in kinetic energy is called as work done. Work done = (K. E. )in – ( K. E. )fin Work done is a scalar quantity thus it has only magnitude

b) Given, Mass = 400 kg, Initial velocity = u = 32 m/s Final velocity = v = 0 m/s.

As,

Work done = ( K. E. )in – ( K. E. )fin

Work done = ½ mv² – ½ mu²

Work done = ½ m (v² – u²)

Work done = ½ × 400 ×( 0²- 32²)

Work done = 200 ×(- 1024)

Work done = -204800 J.

Work done = -204.8 KJ

Minus sign indicates that direction of force and displacement is opposite.

The value of work done will be -204.8 KJ.

**2.) a) Total energy of the system remains constant. Explain this statement in your own words.**

**b) A bus of mass 2540 kg is moving with 45 km/hr. Calculate Kinetic energy associated with it. **

Answer:

a) The law of conservation of energy states that, the total energy of an isolated system is always conserved. We can not create energy or not be destroyed. The energy is converted into one form to another.

Example- Consider a stone is at ground. It contains zero total energy. But someone do work and bring that stone at maximum height then the workdone is stored into energy. Thus at maximum height, the stone contains potential energy only.

When thus stone is released from top then there will increase in kinetic energy and decrease in potential energy. Thus total energy of the system remains constant.

b) Given, Mass = 2540 kg,

Velocity = 45 km/hr

Velocity = 45 × 5/18

Velocity = 12.5 m/s.

As we know that,

Kinetic energy = ½ mv²

Kinetic energy = ½ × 2540 × 12.5

Kinetic energy = 1270 ×12.5

Kinetic energy = 15875 J.

Kinetic energy = 15.9 KJ

**3.) a) Explain potential energy in your own words.**

**b) If the object is placed at height of 8 m then it contains 120 J of energy. Calculate its potential energy If the object is at 14 m. (Take g = 10 m/s²)**

Answer:

**a) Potential energy-**

- Potential energy is a scalar quantity.
- Potential energy always depend on mass and displacement of the object.
- The SI unit of potential energy is Joule.
- Gravitational potential energy depends on the value of acceleration due to gravity, Mass of the object and height.
- It is calculated by using,

Potential energy = mgh

- Potential energy also depend on shape of the object.
- The stretched string, stretched rubber band, elastic stretched spring are the examples of potential energy. The value of potential energy is depend on displacement from mean position.

b) Given, height = 8 m, g = 10 m/s²,

The energy contained in the object because of height is called potential energy.

Potential energy = 120 J.

So firstly we find mass the object.

Potential energy = mgh

m = potential energy /gh

m = 120/ (10× 8)

m = 1.5 Kg.

The mass of the object will be 1.5 kg.

Now we calculate energy if the object is placed at 14 m.

As,

Potential energy = mgh

Potential energy = 1.5 × 10 × 14

Potential energy = 210 J.

The potential energy contained in the stone will be 210 J.

**4.) Ram and Shyam climb a tree of height 80 m. The mass of ram is 56 kg while the mass of Shyam is 74 kg. Ram climbs tree in 35 seconds while Shyam climbs it in 56 seconds. Calculate power of each.**

Answer:

Firstly we calculate power exerted by Ram.

Given, mass of Ram = 56 kg, Height of tree = 80 m,

g = 9.8 m/s², time required to climb try by Ram = 35 seconds.

As we know that,

Power = Potential energy/time

Power = mgh/t

Power = 56 × 9.8 × 80/ 35

Power =43904/35

Power = 1254.4 Watt

The power delivered by Ram is 1254.4 Watt.

Now we calculate power delivered by Shyam.

Given, mass of Ram = 74 kg, Height of tree = 80 m,

g = 9.8 m/s², time required to climb try by Ram = 56 seconds.

As we know that,

Power = Potential energy/time

Power = mgh/t

Power = 74 × 9.8 × 80/ 56

Power = 58016/56

Power = 1036 Watt

The power delivered by Shyam is 1036 Watt.

**5.) Look at the following figure and calculate work done in each case.**

Answer:

a) Look at figure 1. The angle between force and displacement is 90⁰. Thus we know that, if the angle between force and displacement is 90⁰ then workdone becomes zero.

So in this case value of workdone becomes zero.

b) Look at figure 2.

Force = 3 N.

Displacement = 28 m.

The direction of force and displacement is opposite. Thus the workdone becomes negative and maximum.

Work done = -force × displacement

Work done = -3 × 28

Work done = -84 J.

Thus the value of work done will be – 84 J.

**c) **Look at figure 3.

Force = 3 N.

Displacement = 28 m.

The direction of force and displacement is same. Thus the workdone becomes negative and maximum.

Work done = force × displacement

Work done = 3 × 28

Work done = 84 J.

Thus the value of work done will be – 84 J.

**In case you are missed :- Next Chapter Extra Questions**